$A$ proton of mass $m$ and charge $q$ is moving in a plane with kinetic energy $E$. If there exists a uniform magnetic field $B$,perpendicular to the plane of the motion,the proton will move in a circular path of radius

If a charged particle moves in a gravity-free space with uniform velocity,then which of the following is not possible? ($\overrightarrow{E} =$ electric field,$\overrightarrow{B} =$ magnetic field)

An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

$A$ particle of mass $m$ and charge $q$ enters a region of magnetic field (as shown) with speed $v$. There is a region in which the magnetic field is absent,as shown. The particle after entering the region collides elastically with a rigid wall. The time after which the velocity of the particle becomes anti-parallel to its initial velocity is

An electron moves through a uniform magnetic field $\vec{B} = B_0 \hat{i} + 2 B_0 \hat{j} \ T$. At a particular instant of time,the velocity of the electron is $\vec{v} = 3 \hat{i} + 5 \hat{j} \ m/s$. If the magnetic force acting on the electron is $\vec{F} = 5e \hat{k} \ N$,where $e$ is the magnitude of the charge of an electron,then the value of $B_0$ is . . . . . . $T$.

Two long parallel conductors $S_{1}$ and $S_{2}$ are separated by a distance $10 \, cm$ and carry currents of $4 \, A$ and $2 \, A$ respectively. The conductors are placed along the $x$-axis in the $X-Y$ plane. There is a point $P$ located between the conductors (as shown in the figure). $A$ charged particle of $3 \pi \, C$ is passing through the point $P$ with velocity $\overrightarrow{v} = (2 \hat{i} + 3 \hat{j}) \, m/s$; where $\hat{i}$ and $\hat{j}$ represent unit vectors along the $x$ and $y$ axes respectively. The force acting on the charged particle is $4 \pi \times 10^{-5} (-x \hat{i} + 2 \hat{j}) \, N$. The value of $x$ is: